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Fibonacci proof by induction

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]F_{1}+F_{3}+F_{2n-1}[/tex]=[tex]F_{2n}[/tex]

    2. Relevant equations



    3. The attempt at a solution
    P(k+1):[tex]F_{2k-1}+F_{2k+1}[/tex]=[tex]F_{2k+2}[/tex]
     
  2. jcsd
  3. Dec 7, 2008 #2
    What happened to F_1 and F_3?
     
  4. Dec 7, 2008 #3
    well, it's F_1+F_3+....+F_2n-1
     
  5. Dec 7, 2008 #4
    Then P(k + 1) needs to be changed accordingly.

    What have you tried so far?
     
  6. Dec 7, 2008 #5
    I've tried plugging in numbers.
     
  7. Dec 7, 2008 #6
    And how did that work out?

    The definition of Fibonacci numbers will be helpful in the induction proof.
     
  8. Dec 8, 2008 #7
    If I plug in 1, I just get F_1, so 1=1
    If I plug in 2, I get F_3, so 1+2=F_4, 3=3
     
  9. Dec 8, 2008 #8
    F-1+F-3+...+F_2k-1+F_2k+1
    P(k)+F_2k+1
    F_2k+F_2k+1

    Now I'm stumped...
     
  10. Dec 8, 2008 #9

    HallsofIvy

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    What was "F_1+ F_3+ ...+ F_2n-1"?

    In your first post you said the problem was to prove that
    [tex]F_{1}+F_{3}+F_{2n-1}[/tex]=[tex]F_{2n}[/tex]

    Are you saying now it is actually to prove that
    [tex]F_{1}+F_{3}+\cdot\cdot\cdot +F_{2n-1}[/tex]=[tex]F_{2n}[/tex]?
     
  11. Dec 8, 2008 #10
    Very close. What happens when 2 consecutive Fibonacci numbers are added?
     
  12. Dec 8, 2008 #11
    It equals the 3rd Fibonnacci number.
    F_1+F-2=F_3

    so F_2k+F_2k+1=F_2k+1+1
     
  13. Dec 8, 2008 #12
    And that's exactly what you wanted to show.
     
  14. Dec 8, 2008 #13
    Ok thanks!
     
  15. Dec 9, 2008 #14

    epenguin

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    Sorry but maybe the problem is not properly stated? Are the F defined to be Fibonacci numbers?

    Then F2n = F2n-1 + F2n-2

    So if you are then asking also that

    F2n = F2n-1 + F1 + F3

    then

    F2n-2 = F1 + F3

    which is not making much sense.
     
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